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        riddles >> medium >> Painted Cubes
        
(Message started by: THUDandBLUNDER on Jan 2^nd, 2004, 5:01am)

Title: Painted Cubes
Post by THUDandBLUNDER on Jan 2^nd, 2004, 5:01am

Some unit cubes are stuck together to form a larger cube.
Some of the faces of the larger cube are then painted at random.
When the cube is taken apart it is found that 217 of the unit cubes have paint on them.

How many unit cubes are there?

Title: Re: Painted Cubes
Post by LZJ on Jan 2^nd, 2004, 7:58am

[hide]729 cubes, 3 sides painted[/hide]

Title: Re: Painted Cubes
Post by THUDandBLUNDER on Jan 2^nd, 2004, 2:47pm

Quote:

...3 sides painted

Which 3 sides?

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 2^nd, 2004, 3:30pm

Quote:

Which 3 sides?

::[hide]
Adjacent faces i.e each face adjacent to every other face.
3*(n-1)²+3*(n-1)-216=0
[/hide]::

Title: Re: Painted Cubes
Post by Sir Col on Jan 3^rd, 2004, 6:01pm

An unknown number of faces of a cube, n³, are painted. Find the set of values, C, representing the number of cubes that will have paint on them.

For 2³, C={0, 4, 6, 7, 8}.
For 3³, C={0, 9, 15, 18, 19, 21, 23, 24, 25, 26}

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 3^rd, 2004, 9:54pm

::[hide]
for n³,
C={0, n², 2n², 2n²-n, 3n²-2n, 3n²-3n+1, 4n²-5n+2, 5n²-8n+4, 6n²-12n+8}

what!! did i miss something?? please fill in them will ya! ;)
[/hide]::

Title: Re: Painted Cubes
Post by LZJ on Jan 3^rd, 2004, 11:05pm

You missed out [hide] 4n² - 4n [/hide] :)

Title: Re: Painted Cubes
Post by THUDandBLUNDER on Jan 4^th, 2004, 2:51am

For each of the 10 configurations, how many ways are there to paint them?

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 4^th, 2004, 4:51am

on 01/03/04 at 23:05:55, LZJ wrote:

You missed out [hidden] :)

Thx m8!!!

on 01/04/04 at 02:51:27, THUDandBLUNDER wrote:

For each of the 10 configurations, how many ways are there to paint them?

::[hide]
0 face : C(6,0)
1 face : C(6,1)
2 face : C(6,1)*C(1,1)
2 face : C(6,1)*C(4,1)
3 face : C(6,1)*C(4,1)*C(2,1)
3 face : C(6,1)*C(4,1)*C(2,1)
4 face : C(6,1)*C(4,1)*C(2,1)*C(1,1)
4 face : C(6,1)*C(4,1)*C(2,1)*(C(2,1)+C(3,1))
5 face : C(6,5)
6 face : C(6,6)
[/hide]::
i am just finished with my exams, so chances that i have seriously goofed up in the above solution is more than 50%.

Title: Re: Painted Cubes
Post by THUDandBLUNDER on Jan 4^th, 2004, 5:35am

TenaliRaman, congratulations on finishing your exams. 8) >:(

Shouldn't the answers for 2 faces = the answers for 4 faces?

It's probably easier to do it by inspection (of a die).

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 4^th, 2004, 9:39am

on 01/04/04 at 05:35:03, THUDandBLUNDER wrote:

TenaliRaman, congratulations on finishing your exams. 8)

thanks!!! :)
but i would also like to have some luck!!! :D

Quote:

Shouldn't the answers for 2 faces = the answers for 4 faces?

yeah i guess so!! :-[quote]
It's probably easier to do it by inspection (of a die).
[/quote]
that's it!! i don't have a die!! :(
though i will recheck on my calcs..

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 8^th, 2004, 10:56pm

My answers to 0,1,3,5,6 are correct it "seems"!
The answers to 2 and 4 are same and should be equal to the answer i gave for 2 it "seems"!
But it all "seems" doubty!!

Title: Re: Painted Cubes
Post by THUDandBLUNDER on Jan 9^th, 2004, 12:01am

TenaliRaman, your answers may be correct. But they are not clearly correct.
That is, it is not clear which of your answers is which.
Let's call the numbers 2a, 2o, 3c, 3u, 4a, and 4c, for reasons which I hope are obvious.
Anyway, I think your numbers are too high.
For example, consider the case 2o (two opposite faces).
By inspection, the answer is clearly 3; but it "seems" you have an answer of 6 (or 24).
For 2a (two adjacent faces) I get 12, but you "seem" to get 24 (or 6).
"Seems" like you are forgetting to divide by 2.

:P

Title: Re: Painted Cubes
Post by TenaliRaman on Jan 9^th, 2004, 9:48am

Quote:

"Seems" like you are forgetting to divide by 2.

oops!right!! my configurations get repeated twice!!
For the 2 opposite sides, one side can be chosen out 6 in C(6,1) and the opposite side can be chosen only in 1 way that C(1,1).Hence total number of ways it can be done is C(6,1)*C(1,1). However as it is obvious the total number of ways repeats twice.

For the 2 adjacent sides, one side can be chosen out 6 in C(6,1) and the adjacent side can be chosen in 4 ways that C(4,1).Hence total number of ways it can be done is C(6,1)*C(4,1). However as it is obvious again total number of ways repeats twice.

That's how i got my remaining answers as well. So for the 3 side case, i think we might have to divide my answer by 3!=6 i.e.